A person goes from P to Q at a speed of 20 km/h. Then he goes from Q to R at a speed of q km/h. Finally the person goes from R to S at a speed of r km/h. The distances from P to Q, Q to R and R to S are equal. If the average speed from P to R is $\frac{280}{11}$ km/h, and the average speed from Q to S is $\frac{112}{3}$ km/h, then what is the value of r?

40

Let the distance b/w P and Q = Distance b/w Q and R = Distance b/w R and S = D km

We know that,

Speed = \(\frac{Distance}{Time}\)

Time taken from P to Q = \(\frac{D}{20}\)    ----(1)

Time taken from Q to R = \(\frac{2D × 3}{112}\) 

 = \(\frac{3D }{56}\)    ----(2)

Time taken from P to R = \(\frac{2D ×11}{280}\) 

= \(\frac{11D}{140}\) 

Time taken fro Q to R =

On subtracting equation 2 from equation 1,

= \(\frac{11D}{140}\)  - \(\frac{D}{20}\) 

= \(\frac{11D - 7D}{140}\) 

= \(\frac{4D}{140}\)

= \(\frac{D}{35}\)

Time taken from R to S

= \(\frac{3D }{56}\) - \(\frac{D }{35}\) 

= \(\frac{49D }{1960}\) 

= \(\frac{D }{40}\)

It is given that , Speed of train from R to S = r km/h

According to question,

\(\frac{D }{r}\)= \(\frac{D }{40}\) 

r = 40 km/h